Logarithmic Differentiation — Question 6

Question 6 - Solution

The real domain is $|x|\geq1,\ x\ne-2$. Where the function is nonzero, use absolute values in logarithmic differentiation:

$$\ln|f(x)|=4\ln(x^2+3x+5)+\tfrac12\ln(x^2-1)-5\ln|x+2|-\tfrac13\ln|3x-1|$$

Differentiating gives

$$\boxed{f'(x)=f(x)\left(\frac{4(2x+3)}{x^2+3x+5}+\frac{x}{x^2-1}-\frac5{x+2}-\frac1{3x-1}\right).}$$

This formula holds for $|x|>1,\ x\ne-2$. At $x=\pm1$ there is no finite derivative. The cube root is the real cube root.